Interesting 26

Perfect cubes under $27$ are just $8$ and $1$ , the $4$ th answer claims $27$ the lowest number $1$ less than a cube and $1$ more than a square. Simply showing $6=2*3$ is not a perfect square is enough to show this as the right answer

EDIT:

The other solutions are incorrect because:

All numbers of the form $n=pq$ or $n=p^{3}$ for primes $p$ and \q) have $4$ factors, taking $p=3$ and $q=5$ yields $n=15$ has $4$ factors and also has two digits meaning $26$ is not the only two-digit number with $4$ factors

Taking the two smallest square numbers and adding them together will give the smallest sum of two positive perfect squares. $1^{2}+2^{2}=5$ so $26$ is not the smallest number with these qualifications

Reversing the digits of $26$ is $62$ , as three times its digits sum will be the same for these numbers and $62>26$ we see $62$ will also be greater than three times its digits sum. Because $62>26$ , $26$ is not the largest number to meet this qualification